Let $$\alpha$$, $$\beta$$ be the roots of the equation $$x^2 - \sqrt{2}x + 2 = 0$$. Then $$\alpha^{14} + \beta^{14}$$ is equal to

The roots of $$x^2 - \sqrt{2}x + 2 = 0$$:

$$x = \frac{\sqrt{2} \pm \sqrt{2 - 8}}{2} = \frac{\sqrt{2} \pm \sqrt{-6}}{2} = \frac{\sqrt{2} \pm i\sqrt{6}}{2}$$

Converting to polar form: $$|z| = \sqrt{\frac{2 + 6}{4}} = \sqrt{2}$$

So $$\alpha = \sqrt{2}e^{i\theta}$$ and $$\beta = \sqrt{2}e^{-i\theta}$$ for some angle $$\theta$$.

$$\alpha\beta = 2$$ (from the equation, product of roots = 2). $$\alpha + \beta = \sqrt{2}$$.

$$\alpha^{14} + \beta^{14}$$: We use recurrence. Let $$S_n = \alpha^n + \beta^n$$.

$$S_n = (\alpha + \beta)S_{n-1} - \alpha\beta \cdot S_{n-2} = \sqrt{2}S_{n-1} - 2S_{n-2}$$

$$S_0 = 2, \quad S_1 = \sqrt{2}$$

$$S_2 = \sqrt{2} \cdot \sqrt{2} - 2 \cdot 2 = 2 - 4 = -2$$

$$S_3 = \sqrt{2}(-2) - 2\sqrt{2} = -2\sqrt{2} - 2\sqrt{2} = -4\sqrt{2}$$

$$S_4 = \sqrt{2}(-4\sqrt{2}) - 2(-2) = -8 + 4 = -4$$

$$S_5 = \sqrt{2}(-4) - 2(-4\sqrt{2}) = -4\sqrt{2} + 8\sqrt{2} = 4\sqrt{2}$$

$$S_6 = \sqrt{2}(4\sqrt{2}) - 2(-4) = 8 + 8 = 16$$

$$S_7 = \sqrt{2}(16) - 2(4\sqrt{2}) = 16\sqrt{2} - 8\sqrt{2} = 8\sqrt{2}$$

$$S_8 = \sqrt{2}(8\sqrt{2}) - 2(16) = 16 - 32 = -16$$

$$S_9 = \sqrt{2}(-16) - 2(8\sqrt{2}) = -16\sqrt{2} - 16\sqrt{2} = -32\sqrt{2}$$

$$S_{10} = \sqrt{2}(-32\sqrt{2}) - 2(-16) = -64 + 32 = -32$$

$$S_{11} = \sqrt{2}(-32) - 2(-32\sqrt{2}) = -32\sqrt{2} + 64\sqrt{2} = 32\sqrt{2}$$

$$S_{12} = \sqrt{2}(32\sqrt{2}) - 2(-32) = 64 + 64 = 128$$

$$S_{13} = \sqrt{2}(128) - 2(32\sqrt{2}) = 128\sqrt{2} - 64\sqrt{2} = 64\sqrt{2}$$

$$S_{14} = \sqrt{2}(64\sqrt{2}) - 2(128) = 128 - 256 = -128$$

Therefore $$\alpha^{14} + \beta^{14} = -128$$.

This matches option 3: $$-128$$.